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256A: ALGEBRAIC GEOMETRY
Contents
§I.1: Affine Varieties 1
§I.2: Projective Varieties 2
§I.3: Morphisms 3
§I.4: Rational Maps 5
§I.5: Nonsingular Varieties 8
§I.6: Nonsingular Curves 13
§I.7: Intersections in Projective Space 3 14
§I (Supplement): Representing families (Lines in P ) 16
§II.1: Sheaves 17
§II.2: Schemes 20
§II.3: First Properties of Schemes 21
§II.4: Separated and Proper Morphisms 24
§II.5: Sheaves of Modules 26
§II.6: Divisors and §II.7: Projective Morphisms 29
The following are notes from a course taught by Robin Hartshorne intending to
cover the first two chapters of his text Algebraic Geometry. Only the supplementary
comments and examples are included.
§I.1: Affine Varieties
Example. If k = R, A = R[x,y], then the variety Z(y − mx − b) defines a variety,
a line.
2 2 2 2 2 2
Note that x + y = 1 gives a circle, but Z(x + y + 1) = ∅ and Z(x + y )
consists of a single point—this is because R = C 6= R.
If k = Q, R, or F , think of the variety as contained in the algebraic closure k
p
and do algebraic geometry in this affine space, then look for points over k.
Example. R1 in its usual topology has dimR1 = 0 since the only irreducible subset
Y is a point: if a 6= b ∈ Y , choose a < c < b, so that R = (−∞,c]∪[c,∞), and one
canintersect this with Y to obtain a decomposition of Y . This works more generally
for any Hausdorff space: for any two points in a subset, find the corresponding open
sets U ∩V = ∅ containing these points and take their complements.
For us, dimension is given by chains of distinct primes.
Definition. There are four equivalent ways to define the dimension of a ring:
Notes by John Voight, jvoight@math.berkeley.edu, taken from a course taught by Robin
Hartshorne, August 28–December 8, 2000.
1
2 256A: ALGEBRAIC GEOMETRY
(1) For any ring R, we have the Krull dimension, which is
dimR=sup{r:p0 (p1 (···(pr ⊂R}
for distinct prime ideals of R.
(2) Let A be a local noetherian ring with maximal ideal m. Then we define
dimA=inf{n:x ,...,x ∈m, A/hx ,...,x i Artin};
1 n 1 n
recall that a ring is Artin if it is of finite length, i.e. there exists an upper
2
bound for the length of chains of ideals (e.g. k[x]/hx i).
(3) For A local, we define
∞
M i i+1 2
gr A= m/m =k⊕m/m ⊕...
m
i=0
ν ν−1
where k = A/m is the residue field. We have dim m /m <∞,denoted
k
φ (ν). Then there exists a polynomial p with rational coefficients such
A A
that for all sufficiently large ν, φ (ν) = P (ν). We set dimA = degP +1.
A A A
(4) For R an integral domain containing a field k, we consider k ⊂ K(R) the
field of fractions of R. Then trdegk K(R) = dimR.
Wehave the following:
Theorem. If R is a finitely generated k-algebra, then trdegk K(R) is equal to the
Krull dimension. If R is any noetherian ring, dimR = sup dimRp. If A is a
p⊂R
local noetherian ring with maximal ideal m, then the definitions above agree with
the Krull dimension.
Wecan compute the dimension of An in many ways.
Example. We note that dimAn = n for k = k. Let A = k[x ,...,x ] so by defi-
k 1 n
nition dimAn = dimA. This follows now immediately since K(A) = k(x ,...,x )
k 1 n
has transcendence degree n over k. Alternatively, we have h0i ⊂ hx i ⊂ ··· ⊂
1
hx ,...,x i so dimA ≥ n. But the localization A when divided by
1 n hx ,x ,...,x i
1 2 n
hx ,...,x i gives k which is Artin, so dimA ≤ n. Finally, gr A = A, and φ(ν)
1 n m
counts the number of monomials of degree ν in x ,...,x , which totals n+ν−1 ,
1 n n−1
which is a polynomial in ν of degree n −1.
§I.2: Projective Varieties
Here is a concrete description of projective space:
1 ×
Example. The projective line P is the set of points (a : a ) modulo k . If a 6= 0,
k 0 1 0
we can take as a representative (a : a ) = (1 : a /a ) = (1 : b) for b ∈ k; if a = 0,
0 1 1 0 0
a 6= 0 by definition so (a : a ) = (0 : a ) = (0 : 1). Therefore as a set,
1 0 1 1
1 1
P ={(1:b):b∈k}∪{(0:1)}=A ∪{∞}.
k k
2 2 1
Similarly, the projective plane is P = A ∪P , including the line at infinity.
k k k
Looking at projective versions of affine varities can lead to some very important
(and surprising) information:
2 2 2
Example. We have A ≃ U ⊂ P where U = P \Z(x ). Therefore the conic
R 0 R 0 R 0
2 2 2 2 2
x +x =x is of the form x +y = 1 and does not intersect the line at infinity
1 2 0
(as is plain from the graph).
2 2
Alternatively, the curve C : y = x lifts to C : x x = x , so x = 0 implies
0 2 1 0
2
x =0, and we have the single intersection point (0 : 0 : 1). Looking in U , we see
1 2
that the parabola is tangent to the line at infinity.
256A: ALGEBRAIC GEOMETRY 3
3 2 3
Finally, the seemingly honest curve y = x has the projective closure x x = x ,
0 2 3
which looks on the set x 6= 0 like u2 = v3, so the curve has a cusp at infinity!
2
Here is an extended description of the twisted cubic curve.
2 3
Example. The affine version of the twisted cubic curve C is the subset {(a,a ,a ) :
a ∈ k} ⊂ A3, i.e. the set of points parameterized by x = t, y = t2, z = t3.
k
Claim. C is a closed, irreducible subset of dim1.
To see this, we find the prime ideal p ⊂ k[x,y,z] = A defining this ideal. We
ψ 2 3
map A −→ k[t] by x 7→ t, y 7→ t , z 7→ t ; since the image is a domain, kerψ = p
is prime. We guess that p = I(C). If f = f(x,y,z) ∈ p, then f vanishes on C:
2 3
f(a,a ,a ) = ψ(f)(a) = 0, so C ⊂ Z(p). Conversely, if P = (a,b,c) ∈ Z(p),
2 3 2 3
then for all f ∈ p, f(a,b,c) = 0. Since y − x , z − x ∈ p, b = a , c = a , so
2 3
P =(a,a ,a ). So C = Z(p), so C is certainly a closed and irreducible subset. In
particular, dimC = dimA/p = dimk[t] = 1, which proves the claim.
2 3
Howmanyequations define C? Take f = y−x , f = z−x . Then Z(f ,f ) =
1 2 1 2
C, since hf ,f i ⊂ p, so Z(f ,f ) ⊃ Z(p) = C, but we have actually shown by the
1 2 1 2
above equality just on these generators. How many equations define the prime p?
Simply, p = hf ,f i since A/hf ,f i = k[x] already.
1 2 1 2
3 3 3 n
Now projectivize C: We have A ≃ U ⊂ P , U = P \Z(x ). Then C ⊂ A ⊂
0 0 0
Pn ⊃ C. For any set V, if V is irreducible, then V is irreducible. Therefore C is a
n
closed irreducible subset of P of dimension 1.
2 2 3
We homogenize p directly and have g = yw − x , g = w z − x ∈ p = I(C).
1 2
Do these equations define C? No, because if x = w = 0, L ⊂ Z(g ,g ), but C
1 2
is not a line and is irreducible. We also have y2 − xz = g3 ∈ p. We would like
C=Z(g1,g2,g3).
Weknowthat C ⊂Z(g ,g ,g ). Next, if P = (a : b : c : d) ∈ C, if P is an affine
1 2 3
point (d 6= 0), then P ∈ C by earlier work. Otherwise, d = 0, so a = 0 and then
b = 0, so P = (0 : 0 : 1 : 0) ∈ P3. For the moment, we will omit the reason why
P ∈C.
Instead, we ask if g1,g2,g3 generate p. If g(x,y,z,w) ∈ p we can substitute for
2 3 2
the x , x , and y terms, so what is left is of the form
h (z,w)+xh (z,w)+yh (z,w)+xyh (z,w).
1 2 3 4
ψ 2 2 3 3
We now take k[x,y,z,w] −→ k[t,u] by x,y,z,w 7→ tu ,t u,t ,u . We find g4 =
xy −zw ∈ p which allows us to remove the h4 term, and under this substitution
the h are cubes (in t and u), so it must be identically zero. Since xg −wg = g ,
i 1 4 2
we have p = hg ,g ,g i, so indeed P ∈ C.
1 3 4
Claim. p cannot be generated by < 3 elements.
p is a homogeneous ideal so S ⊃ p = L∞ pd. We have p0 = 0 and p1 = 0.
d=0
p is the k-vector space generated by the qudratic polynomials g ,g ,g ⊂ S =
2 1 2 4 2
2 2 2 2
k{x ,xy,xz,xw,y ,yz,yw,z ,zw,w }, a space of dimension 10. We must have the
2 2
gi linearly independent over k, because dividing out by z, w, we find x ,y ,xy are
linearly independent.
§I.3: Morphisms
Here are examples of regular functions:
4 256A: ALGEBRAIC GEOMETRY
Example. If we take the affine line A1, an open set V ∋ 0, then f is regular if
k
f = g/h with h(0) 6= 0; h has finitely many zeros, so we can shrink the open set,
and we find f is regular at 0 iff f ∈ k[x] .
hxi
Example. If we take the projective line P1, we find O(V ) = k. For U = P1\Z(x ) =
0 0
1 1 1
A with O(U ) = k[x]. But U ⊂ P \Z(x ) = A with O(U ) = k[y], y = 1/x. A
0 1 1 1
function that is a polynomial in x and 1/x is constant.
Here is an alternative proof of:
Theorem (Theorem 3.2(a)). O(Y) ≃ A(Y) when Y is affine.
Proof. Let f = g /h on the open set U , on any open cover such that V = S U .
i i i p i i
h 6= 0 on U and the U cover Y, so Z(h ,h ,...) ∩ V = ∅, so hh ,h ...i =
i i P i 1 2 1 2
A(V) so 1 = r a h (the sum is finite) for certain a ∈ A(V), and thus f =
i=1 i i i
P P
r a fh = r a g ∈ A(V) since fh = g on U which is dense in V (f is
i=1 i i i=1 i i i i i
continuous).
Let C be the category of varieties, with objects varieties and the arrows mor-
phisms.
1 φ 2
Example. A −→ Y = Z(y) ⊂ A by x 7→ (x,0) has f(x,y)/g(x,y) = f(x,0)/g(x,0)
regular on A1, therefore A1 ≃ Y since ψ : Y → A1 by (x,0) 7→ x is an inverse.
2 2 1 2
Example. The variety Y = Z(y − x ) ⊂ A has Y ≃ A by (x,x ) 7→ x. To see
this, we prove:
∗
Lemma. If V,W are affine varieties, and φ : V → W a morphism, we have φ :
∗
A(W)=O(W)→O(V)=A(V). φ is an isomorphism iff φ is an isomorphism.
∗
Proof. φ an isomorphism implies φ an isomorphism is true for any (not necessarily
affine) V,W. Use correspondence: P ∈ V ↔ m ⊂ A(V). We define ψ : W → V
P
∗
using the equivalence Q ∈ W ↔ m ⊂ A(W), since φ is an isomorphism, and thus
Q
ψisbijective. The map is a homeomorphism because ψ∗ takes ideals to ideals—just
carry over quotients of functions.
∗ 2
Returning to the example, we find φ : k[x,y]/hy−x i → k[x] is an isomorphism,
so φ is an isomorphism.
Here is some more category language. If C is the category of varieties, O is a
contravariant functor from C to k-algebras (domains), since a map V → W induces
a map O(W) → O(V). We have in fact that the subcategory of affine varieties
mapping to the subcategory of finitely generated k-algebras is an equivalence of
categories. For if A is such an algebra, k[x ,...,x ] → A by x 7→ a for generators
1 n i i
a is surjective (if V is a variety defined by p in An, a different choice of a gives an
i i
isomorphic V).
Nowforsomething really wild: If we take the subcategory of C of those varieties
such that O(V) is a finitely generated k-algebra, we can look at the adjoint functor
F. If V is a variety, WA = F(A) finitely generated with a map φ to O(V), if
P ∈ V, we consider m , which is not necessarily a one-to-one map, but we can
P
−1
still have m ⊂ φ (m ) ⊂ A maximal (look at the quotient fields), so we have
Q P
Q∈WA. In other words, HomC(V,F(A)) = Hom(A,O(V)), so the functors are
adjoint. (Indeed, one can define affine varieties in this way.)
Example. It is possible to have φ : V → W that is a bijective homeomorphism but
is not an isomorphism.
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